# Class Exercise #3 1.101 Structures Work ```Class Exercise #3
1.101 Structures Work
The figure shows our simply supported beam. The displace&shy;
ment of the centroidal axis (z=0) was given to you as
L
z,
F
a
w
midspan
c
x
∆ =?
c
a
x
Fa
2
2
= –  ------------ ⋅ (3L – 4a )= –∆
24 EI
L
, P
F
∆ = - w(0)
To deduce an expression fro the midspan displacement of the
beam shown at the left - load P is concentrated at mid-span, we
set a = L/2, and F = P/2. (Think of moving the two loads, each
of magnitude F, in toward the midspan, shrinking c to zero).
Making these substitutions, we have:
3
w
midspan
PL
= –  ----------&shy;-
48EI
For the beam two “flanges” of equal area, A, located at +/- H/2 from the centroid
of the section and with “web” infinitely thin, the second moment of area, with
respect to the centroid, is approximately
b
h A
H/2
2
H
I = 2 ⋅  ---- ⋅ A
2
A more accurate result comes from integrating over the two sections. Doing one
section is all we need do because of symmetry. Thus
z = (H + h) ⁄ 2
I = 2⋅∫
z = (H – h) ⁄ 2
H/2
A
2
z ⋅ bdz which evaluates to
3
3
  H 2
 H2
bh 
bh 
I = 2 ⋅   ---- ⋅ bh + -------- = 2 ⋅  ---- ⋅ A + -------- 
12 
12 
 2
2
The first term within the bracket is just what we had as our approximation. The second term we recognize
is just the “local” second moment of area of the top bit. Our approximation is good if (h/H)2 is small for we
can write:
3
  H 2

bh 
H 2
1 h 2
I = 2 ⋅  ---- ⋅ bh + -------- = 2 ⋅  ---- ⋅ bh  1 + --- ⋅  ---- 
 
 2
12 
3  H 
 2

e.g., if h = 0.1*H then I is less than 1% more than our approximate result.
November 1, 2005
LL Bucciarelli
```